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Transcript
AP Chemistry
Practice Exam
®
From the 2015 Administration
NOTE: This is a modified version of the 2015 AP Chemistry Exam.
This Practice Exam is provided by the College Board for AP Exam
preparation. Teachers are permitted to download the materials and
make copies to use with their students in a classroom setting only.
To maintain the security of this exam, teachers should collect all
materials after their administration and keep them in a secure location.
Exams may not be posted on school or personal websites, nor
electronically redistributed for any reason. Further distribution of
these materials outside of the secure College Board site disadvantages
teachers who rely on uncirculated questions for classroom testing.
Any additional distribution is in violation of the College Board’s
copyright policies and may result in the termination of Practice Exam
access for your school as well as the removal of access to other online
services such as the AP Teacher Community and Online Score Reports.
Contents
Exam Instructions
Student Answer Sheet for the Multiple-Choice Section
Section I: Multiple-Choice Questions
Section II: Free-Response Questions
Multiple-Choice Answer Key
Free-Response Scoring Guidelines
Scoring Worksheet
Note: This publication shows the page numbers that appeared in
the 2014−15 AP Exam Instructions book and in the actual exam.
This publication was not repaginated to begin with page 1.
© 2015 The College Board. College Board, Advanced Placement Program, AP, SAT and the acorn logo are
registered trademarks of the College Board. All other products and services may be trademarks of their
respective owners. Permission to use copyrighted College Board materials may be requested online at:
www.collegeboard.com/inquiry/cbpermit.html.
Exam Instructions
The following contains instructions taken from
the 2014−15 AP Exam Instructions book.
AP® Chemistry Exam
Regularly Scheduled Exam Date: Monday morning, May 4, 2015
Late-Testing Exam Date: Thursday afternoon, May 21, 2015
Section I Total Time: 1 hr. 30 min. Section II Total Time: 1 hr. 45 min.
Section I
Section II
Total Time: 1 hour 30 minutes
Calculator not permitted
Percent of Total Score: 50%
Writing Instrument: Pencil required
Number of Questions: 60*
Total Time: 1 hour 45 minutes
Calculators allowed for all of Section II
Percent of Total Score: 50%
Writing Instrument: Either pencil or pen
with black or dark blue ink
Number of Questions: 7
*The number of questions may vary slightly
depending on the form of the exam.
(3 ten-point and 4 four-point
questions; 105 minutes)
What Proctors Need to Bring to This Exam
•
•
•
•
•
•
•
Exam packets
Answer sheets
AP Student Packs
2014-15 AP Coordinator’s Manual
This book — AP Exam Instructions
AP Exam Seating Chart template(s)
School Code and Home-School/SelfStudy Codes
• Extra calculators
• Pencil sharpener
• Container for students’ electronic
devices (if needed)
• Extra No . 2 pencils with erasers
• Extra pens with black or dark blue ink
• Extra paper
• Stapler
• Watch
• Signs for the door to the testing room
– “Exam in Progress”
– “Cell phones are prohibited in the
testing room”
Note: Tables of equations and constants are provided in the exam booklets for both sections of
the exam .
Students are not allowed to use calculators in Section I of the AP Chemistry Exam . However,
students are permitted to use four-function, scientific, or graphing calculators to answer
questions in Section II . Before starting the exam administration, make sure that each student has
an appropriate calculator and that any student with a graphing calculator has a model from the
approved list on page 46 of the 2014-15 AP Coordinator’s Manual . See pages 43–46 of the
2014-15 AP Coordinator’s Manual for more information . If a student does not have an appropriate
calculator or has a graphing calculator not on the approved list, you may provide one from your
supply . If the student does not want to use the calculator you provide, or does not want to use a
calculator at all, he or she must hand copy, date, and sign the release statement on page 44 of the
2014-15 AP Coordinator’s Manual .
42
© 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org.
AP Exam Instructions
During the administration of Section II students may have no more than two calculators on
their desks . Calculators may not be shared . Calculator memories do not need to be cleared
before or after the exam . Students with Hewlett-Packard 48–50 Series and Casio FX-9860
graphing calculators may use cards designed for use with these calculators . Proctors should make
sure infrared ports (Hewlett-Packard) are not facing each other . Since graphing calculators
can be used to store data, including text, proctors should monitor that students are using
their calculators appropriately. Attempts by students to use the calculator to remove exam
questions and/or answers from the room may result in the cancellation of AP Exam scores.
Students will be allowed to use the table of equations and constants on both sections of the exam .
SECTION I: Multiple Choice
!
Do not begin the exam instructions below until you have completed the appropriate
General Instructions for your group.
Make sure you begin the exam at the designated time . Remember: You must complete a seating
chart for this exam . See pages 279–280 for a seating chart template and instructions . See the
2014-15 AP Coordinator’s Manual for exam seating requirements (pages 48–50, 88) .
If you are giving the regularly scheduled exam, say:
It is Monday morning, May 4, and you will be taking the AP Chemistry Exam.
If you are giving the alternate exam for late testing, say:
It is Thursday afternoon, May 21, and you will be taking the AP Chemistry Exam.
In a moment, you will open the packet that contains your exam materials.
By opening this packet, you agree to all of the AP Program’s policies and
procedures outlined in the 2014 -15 Bulletin for AP Students and Parents. You
may now remove the shrinkwrap from your exam packet and take out the
Section I booklet, but do not open the booklet or the shrinkwrapped Section II
materials. Put the white seals aside. . . .
Carefully remove the AP Exam label found near the top left of your exam
booklet cover. Now place it on page 1 of your answer sheet on the light blue
box near the top right-hand corner that reads “AP Exam Label.”
Read the statements on the front cover of Section I and look up
when you have finished. . . .
Sign your name, and write today’s date. Look up when you
have finished. . . .
Now print your full legal name where indicated. Are there any questions? . . .
Turn to the back cover and read it completely. Look up when you
have finished. . . .
43
© 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org.
CHEMISTRY
If students accidentally place the exam label in the space for the number label or vice versa, advise
them to leave the labels in place . They should not try to remove the label; their exam will be
processed correctly .
Chemistry
Are there any questions? . . .
You will now take the multiple-choice portion of the exam. You should have
in front of you the multiple-choice booklet and your answer sheet. You may
never discuss these specific multiple-choice questions at any time in any form
with anyone, including your teacher and other students. If you disclose these
questions through any means, your AP Exam score will be canceled. . . .
You must complete the answer sheet using a No. 2 pencil only. Mark all of
your responses beginning on page 2 of your answer sheet, one response per
question. Completely fill in the circles. If you need to erase, do so carefully
and completely. No credit will be given for anything written in the exam
booklet. Scratch paper is not allowed, but you may use the margins or
any blank space in the exam booklet for scratch work. Calculators are not
allowed for this section. Please put your calculators under your chair. Are
there any questions? . . .
You have 1 hour and 30 minutes for this section. Open your Section I booklet
and begin.
12
. Note Stop Time here
. Check that students are
Note Start Time here
marking their answers in pencil on their answer sheets, and that they are not looking at their
shrinkwrapped Section II booklets . After 1 hour and 20 minutes, say:
9
3
6
There are 10 minutes remaining.
After 10 minutes, say:
Stop working. Close your booklet and put your answer sheet on your desk,
face up. Make sure you have your AP number label and an AP Exam label on
page 1 of your answer sheet. Sit quietly while I collect your answer sheets.
Collect an answer sheet from each student . Check that each answer sheet has an AP number label
and an AP Exam label . After all answer sheets have been collected, say:
Now you must seal your exam booklet using the white seals you set aside
earlier. Remove the white seals from the backing and press one on each area
of your exam booklet cover marked “PLACE SEAL HERE.” Fold each seal over
the back cover. When you have finished, place the booklet on your desk, face
up. I will now collect your Section I booklet. . . .
Collect a Section I booklet from each student . Check that each student has signed the front cover
of the sealed Section I booklet .
There is a 10-minute break between Sections I and II . When all Section I materials have been
collected and accounted for and you are ready for the break, say:
Please listen carefully to these instructions before we take a 10-minute
break. All items you placed under your chair at the beginning of this exam
must stay there, and you are not permitted to open or access them in
any way. Leave your shrinkwrapped Section II packet on top of your desk
during the break. You are not allowed to consult teachers, other students,
or textbooks during the break. You may not make phone calls, send text
messages, check email, use a social networking site, or access any electronic
44
© 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org.
AP Exam Instructions
or communication device. Remember, you may never discuss the multiplechoice questions at any time in any form with anyone, including your teacher
and other students. If you disclose these questions through any means, your
AP Exam score will be canceled. Are there any questions? . . .
12
9
6
3
You may begin your break. Testing will resume at
.
SECTION II: Free Response
After the break, say:
May I have everyone’s attention? Place your Student Pack on your desk. . . .
You may now remove the shrinkwrap from the Section II packet, but do not
open the exam booklet until you are told to do so. . . .
Read the bulleted statements on the front cover of the exam booklet. Look
up when you have finished. . . .
Now place an AP number label on the shaded box. If you don’t have any
AP number labels, write your AP number in the box. Look up when you
have finished. . . .
Read the last statement. . . .
Using your pen, print the first, middle and last initials of your legal name
in the boxes and print today’s date where indicated. This constitutes your
signature and your agreement to the statements on the front cover. . . .
Turn to the back cover and complete Item 1 under “Important Identification
Information.” Print the first two letters of your last name and the first letter
of your first name in the boxes. Look up when you have finished. . . .
In Item 2, print your date of birth in the boxes. . . .
In Item 3, write the school code you printed on the front of your Student
Pack in the boxes. . . .
Read Item 4. . . .
Are there any questions? . . .
While Student Packs are being collected, read the information on the back
cover of the exam booklet. Do not open the exam booklet until you are told to
do so. Look up when you have finished. . . .
Collect the Student Packs . Then say:
Are there any questions? . . .
45
© 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org.
CHEMISTRY
I need to collect the Student Pack from anyone who will be taking another
AP Exam. You may keep it only if you are not taking any other AP Exams this
year. If you have no other AP Exams to take, place your Student Pack under
your chair now. . . .
Chemistry
Calculators may be used for Section II. You may get your calculators from
under your chair and place them on your desk. . . .
You have 1 hour and 45 minutes to complete Section II. You are responsible
for pacing yourself, and you may proceed freely from one question to the
next. You must write your answers in the exam booklet using a pen with
black or dark blue ink or a No. 2 pencil. If you use a pencil, be sure that your
writing is dark enough to be easily read. If you need more paper during the
exam, raise your hand. At the top of each extra sheet of paper you use, be
sure to write only your AP number and the number of the question you are
working on. Do not write your name. Are there any questions? . . .
You may begin.
12
. Note Stop Time here
. Proctors should also
Note Start Time here
make sure that Hewlett-Packard calculators’ infrared ports are not facing each other and that
students are not sharing calculators . After 1 hour and 35 minutes, say:
9
3
6
There are 10 minutes remaining.
After 10 minutes, say:
Stop working and close your exam booklet. Place it on your desk, face up. . . .
If any students used extra paper for the free-response section, have those students staple the extra
sheet(s) to the first page corresponding to that question in their exam booklets . Complete an
Incident Report and include any exam booklets with extra sheets of paper in an Incident Report
return envelope (see page 57 of the AP Coordinator’s Manual for details) . Then say:
Remain in your seat, without talking, while the exam materials
are collected. . . .
Collect a Section II booklet from each student . Check for the following:
• Exam booklet front cover: The student placed an AP number label on the shaded box, and
printed his or her initials and today’s date .
• Exam booklet back cover: The student completed the “Important Identification
Information” area .
When all exam materials have been collected and accounted for, return to students any electronic
devices you may have collected before the start of the exam .
If you are giving the regularly scheduled exam, say:
You may not discuss or share these specific free-response questions with anyone
unless they are released on the College Board website in about two days. Your AP
Exam score results will be available online in July.
If you are giving the alternate exam for late testing, say:
None of the questions in this exam may ever be discussed or shared in any way
at any time. Your AP Exam score results will be available online in July.
46
© 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org.
AP Exam Instructions
If any students completed the AP number card at the beginning of this exam, say:
Please remember to take your AP number card with you. You will need the
information on this card to view your scores and order AP score reporting
services online.
Then say:
You are now dismissed.
All exam materials must be placed in secure storage until they are returned to the AP Program
after your school’s last administration . Before storing materials, check the “School Use Only”
section on page 1 of the answer sheet and:
• Fill in the appropriate section number circle in order to access a separate AP
Instructional Planning Report (for regularly scheduled exams only) or subject
score roster at the class section or teacher level . See “Post-Exam Activities” in the
2014-15 AP Coordinator’s Manual .
• Check your list of students who are eligible for fee reductions and fill in the
appropriate circle on their registration answer sheets .
Be sure to give the completed seating chart to the AP Coordinator . Schools must retain seating
charts for at least six months (unless the state or district requires that they be retained for a longer
period of time) . Schools should not return any seating charts in their exam shipments unless they
are required as part of an Incident Report .
CHEMISTRY
47
© 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org.
Student Answer Sheet for
the Multiple-Choice Section
Use this section to capture student responses. (Note that the following
answer sheet is a sample, and may differ from one used in an actual exam.)
(from Student Pack)
AP Number Label
USE NO. 2 PENCIL ONLY
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H. AP EXAM I AM
TAKING USING THIS
ANSWER SHEET
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Section Number
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Date
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Omit apostrophes, Jr., II.
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A. SIGNATURE
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L. SOCIAL SECURITY
NUMBER (Optional)
INTERNATIONAL PHONE
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I. AREA CODE AND
PHONE NUMBER
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C. YOUR AP NUMBER
B123456789T
To maintain the security of the exam and the validity of my AP score, I will allow no one else to see the multiple-choice questions. I will
seal the multiple-choice booklet when asked to do so, and I will not discuss these questions with anyone at any time after completing the
section. I am aware of and agree to the AP Program’s policies and procedures as outlined in the 2014-15 Bulletin for AP Students and
Parents, including using testing accommodations (e.g., extended time, computer, etc.) only if I have been preapproved by College Board
Services for Students with Disabilities.
COMPLETE THIS AREA AT EVERY EXAM.
Answer Sheet
2015
104246-00657• UNLWEB315
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M. COLLEGE TO RECEIVE YOUR
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SERIAL NUMBER
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TIME
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O. SURVEY QUESTIONS — Answer the survey questions in the AP Student Pack. Do not put responses to exam questions in this section.
1
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P. LANGUAGE — Do not complete this section unless instructed to do so.
If this answer sheet is for the French Language and Culture, German Language and Culture, Italian Language and Culture, Spanish Language
and Culture, or Spanish Literature and Culture Exam, please answer the following questions. Your responses will not affect your score.
2. Do you regularly speak or hear the language at home?
1. Have you lived or studied for one month or more in a country where the language of the
exam you are now taking is spoken?
Yes
No
Yes
No
QUESTIONS 1–75
Indicate your answers to the exam questions in this section (pages 2 and 3). Mark only one response per question
for Questions 1 through 120. If a question has only four answer options, do not mark option E. Answers written in
the multiple-choice booklet will not be scored.
EXAMPLES OF
INCOMPLETE MARKS
COMPLETE MARK
1
A
B
C
D
A
B
C
D
A
B
C
D
You must use a No. 2 pencil and marks must be complete. Do not use a mechanical pencil. It
is very important that you fill in the entire circle darkly and completely. If you change your response,
erase as completely as possible. Incomplete marks or erasures may affect your score.
E
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PT02
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PT04
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DO NOT WRITE IN THIS AREA
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PAGE 3
QUESTIONS 76–120
Be sure each mark is dark and completely fills the circle. If a question has only four answer options, do not mark option E.
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QUESTIONS 121–126
For Students Taking AP Biology
Write your answer in the boxes at the top of the griddable area and fill in the corresponding circles.
Mark only one circle in any column. You will receive credit only if the circles are filled in correctly.
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QUESTIONS 131–142
For Students Taking AP Physics 1 or AP Physics 2
Mark two responses per question. You will receive credit only if both correct responses are selected.
131
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© 2014 The College Board. College Board, AP, Student Search Service and the acorn logo are registered trademarks of the College Board.
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PAGE 4
Section I: Multiple-Choice Questions
This is the multiple-choice section of the 2015 AP exam.
It includes cover material and other administrative instructions
to help familiarize students with the mechanics of the exam.
(Note that future exams may differ in look from the following content.)
For purposes of test security and/or statistical analysis, some
questions have been removed from the version of the exam
that was administered in 2015. Therefore, the timing indicated
here may not be appropriate for a practice exam.
AP Chemistry Exam
®
2015
SECTION I: Multiple Choice
DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.
At a Glance
Total Time
1 hour, 30 minutes
Number of Questions
50
Percent of Total Score
50%
Writing Instrument
Pencil required
Electronic Device
None allowed
Instructions
Section I of this exam contains 50 multiple-choice questions. Fill in only the circles for
numbers 1 through 50 on your answer sheet. Pages containing a periodic table and lists
containing equations and constants are also printed in this booklet.
Indicate all of your answers to the multiple-choice questions on the answer sheet. No
credit will be given for anything written in this exam booklet, but you may use the booklet
for notes or scratch work. After you have decided which of the suggested answers is best,
completely fill in the corresponding circle on the answer sheet.
Because this section offers only four answer options for each question, do not mark the
(E) answer circle for any question. Give only one answer to each question. If you change
an answer, be sure that the previous mark is erased completely. Here is a sample question
and answer.
Use your time effectively, working as quickly as you can without losing accuracy. Do not
spend too much time on any one question. Go on to other questions and come back to
the ones you have not answered if you have time. It is not expected that everyone will
know the answers to all of the multiple-choice questions.
Your total score on Section I is based only on the number of questions answered correctly.
Points are not deducted for incorrect answers or unanswered questions.
Form I
Form Code 4LBP4-S
25
-2-
®
AP CHEMISTRY EQUATIONS AND CONSTANTS
Throughout the exam the following symbols have the definitions specified unless otherwise noted.
L, mL
g
nm
atm
=
=
=
=
liter(s), milliliter(s)
gram(s)
nanometer(s)
atmosphere(s)
mm Hg
J, kJ
V
mol
=
=
=
=
ATOMIC STRUCTURE
millimeters of mercury
joule(s), kilojoule(s)
volt(s)
mole(s)
E = energy
 = frequency
 = wavelength
E = h
c = 
Planck’s constant, h = 6.626  1034 J s
Speed of light, c = 2.998  108 m s1
Avogadro’s number = 6.022  1023 mol1
Electron charge, e = 1.602 × 1019 coulomb
EQUILIBRIUM
Kc =
Kp =
[C]c [D]d
, where a A + b B  c C + d D
[A]a [B]b
Equilibrium Constants
Kc
Kp
Ka
Kb
Kw
(PC )c (PD )d
(PA )a (PB )b
+
Ka = [H ][A ]
[HA]
+
Kb = [OH ][HB ]
(molar concentrations)
(gas pressures)
(weak acid)
(weak base)
(water)
[B]
Kw = [H ][OH] = 1.0  1014 at 25C
= Ka  K b
+
pH = log[H+] , pOH = log[OH]
14 = pH + pOH
pH = pKa + log [A ]
[HA]
pKa = logKa , pKb = logKb
KINETICS
ln[A] t  ln[A] 0 =  kt
1 - 1
[A ]t
[A ]0
k = rate constant
t = time
t ½ = half-life
= kt
t ½ = 0.693
k
-3-
GASES, LIQUIDS, AND SOLUTIONS
PV = nRT
PA = Ptotal  XA, where XA =
moles A
total moles
Ptotal
= PA + PB + PC + . . .
n = m
M
K = C + 273
D= m
V
KE per molecule = 1 mv 2
P
V
T
n
m
M
D
KE
v
A
a
b
c
=
=
=
=
=
=
=
=
=
=
=
=
=
pressure
volume
temperature
number of moles
mass
molar mass
density
kinetic energy
velocity
absorbance
molar absorptivity
path length
concentration
2
Molarity, M = moles of solute per liter of solution
Gas constant, R = 8.314 J mol1 K1
= 0.08206 L atm mol1 K1
A = abc
= 62.36 L torr mol1 K1
1 atm = 760 mm Hg = 760 torr
STP = 273.15 K and 1.0 atm
Ideal gas at STP = 22.4 L mol−1
THERMODYNAMICS / ELECTROCHEMISTRY
q =
m=
c =
T=
S =
H =
G =
n =
E =
I =
q =
t =
q = mcT
 S products -  S reactants
H = Â DHf products - Â DHf reactants
S =
G =
 DGf products -  DGf reactants
G = H  TS
= RT ln K
=  nF E
I =
q
t
heat
mass
specific heat capacity
temperature
standard entropy
standard enthalpy
standard Gibbs free energy
number of moles
standard reduction potential
current (amperes)
charge (coulombs)
time (seconds)
Faraday’s constant, F = 96,485 coulombs per mole
of electrons
1 joule
1 volt =
1 coulomb
-4-
CHEMISTRY
Section I
50 Questions
Time—90 minutes
CALCULATORS ARE NOT ALLOWED FOR SECTION I. Note: For all questions, assume that the temperature is 298 K, the pressure is 1.0 atm, and solutions are aqueous
unless otherwise specified.
Directions: Each of the questions or incomplete statements below is followed by four suggested answers or
completions. Select the one that is best in each case and then fill in the corresponding circle on the answer sheet.
1. A 0.5 mol sample of He(g) and a 0.5 mol sample
of Ne(g) are placed separately in two 10.0 L rigid
containers at 25°C. Each container has a pinhole
opening. Which of the gases, He(g) or Ne(g),
will escape faster through the pinhole and why?
2. The lattice energy of a salt is related to the energy
required to separate the ions. For which of the
following pairs of ions is the energy that is
required to separate the ions largest? (Assume that
the distance between the ions in each pair is equal
to the sum of the ionic radii.)
(A) He(g) will escape faster because the
He(g) atoms are moving at a higher
average speed than the Ne(g) atoms. (B) Ne(g) will escape faster because its initial
pressure in the container is higher.
(C) Ne(g) will escape faster because the
Ne(g) atoms have a higher average kinetic energy than the He(g) atoms. (D) Both gases will escape at the same rate because the atoms of both gases have the same average kinetic energy. (A)
(B)
(C)
(D)
Unauthorized copying or reuse of
any part of this page is illegal.
Na+(g) and Cl−(g) Cs+(g) and Br−(g)
Mg2+(g) and O2−(g)
Ca2+(g) and O2−(g)
GO ON TO THE NEXT PAGE.
-5-
4. Which of the following diagrams best depicts an
alloy of Ni and B ?
(A)
(B) 3. The mass spectrum of element X is presented
in the diagram above. Based on the spectrum,
which of the following can be concluded about
element X?
(C) (A) X is a transition metal, and each peak
represents an oxidation state of the metal.
(B) X contains five electron sublevels.
(C) The atomic mass of X is 90.
(D) The atomic mass of X is between 90 and 92.
(D) Unauthorized copying or reuse of
any part of this page is illegal.
GO ON TO THE NEXT PAGE.
-6-
6. A hot iron ball is dropped into a 200. g sample of
water initially at 50.° C. If 8.4 kJ of heat is
transferred from the ball to the water, what is the
final temperature of the water? (The specific heat
of water is 4.2 J/(g·°C).)
(A)
(B)
(C)
(D)
5. Which of the following is the strongest type of
interaction that occurs between the atoms within
the circled areas of the two molecules represented
above?
(A)
(B)
(C)
(D)
40.°C
51°C
60.°C
70.°C
Polar covalent bond
Nonpolar covalent bond
Hydrogen bond
London dispersion forces
Unauthorized copying or reuse of
any part of this page is illegal.
GO ON TO THE NEXT PAGE.
-7-
Questions 7-9 refer to the following information.
At 27°C, five identical rigid 2.0 L vessels are filled with N2(g) and sealed. Four of the five vessels also contain
a 0.050 mol sample of NaHCO3 (s), NaBr(s), Cu(s), or I2 (s), as shown in the diagram above. The volume taken up
by the solids is negligible, and the initial pressure of N2(g) in each vessel is 720 mm Hg. All four vessels are heated
to 127°C and allowed to reach a constant pressure.
7. At 127°C, the pressure in vessel 1 is found to be higher than that in vessel 2. Which of the following reactions best accounts for the observation? (A) NaHCO3 ( s) Æ Na( s) + HCO3 ( s)
(B) NaHCO3 ( s) Æ NaH( s) + CO3 ( s)
(C) 2 NaHCO3 ( s) Æ Na 2 CO3 ( s ) + CO 2 ( g) + H 2 O( g)
(D) 2 NaHCO3 (s ) + N 2 (g) Æ 2 NaNO3 (s) + C2 H 2 ( g)
Unauthorized copying or reuse of
any part of this page is illegal.
GO ON TO THE NEXT PAGE.
-8-
8. At 127°C, the entire sample of I2 is observed to
have vaporized. How does the mass of vessel 5
at 127°C compare to its mass at 27°C?
(A) The mass is less, since the I2 is in the vapor
phase.
(B) The mass is the same, since the number of
each type of atom in the vessel is constant.
(C) The mass is greater, since the I2 will react
with N2 to form NI3 , which has a greater
molar mass.
(D) The mass is greater, since the pressure is
greater and the particles have a higher
average kinetic energy.
9. The gas particles in vessel 3 at 27°C are
represented in the diagram above. The lengths
of the arrows represent the speeds of the particles.
Which of the following diagrams best represents
the particles when vessel 3 is heated to 127°C?
(A)
(B)
(C)
(D) Unauthorized copying or reuse of
any part of this page is illegal.
GO ON TO THE NEXT PAGE.
-9-
Ionization Energy
(kJ/mol)
10. An acetate buffer solution is prepared by
combining 50. mL of 0.20 M acetic acid,
HC2H3O2(aq), and 50. mL of 0.20 M sodium
acetate, NaC2H3O2(aq). A 5.0 mL sample of
0.10 M NaOH(aq) is added to the buffer solution.
Which of the following is a correct pairing of the
acetate species present in greater concentration
and of the pH of the solution after the NaOH(aq)
is added? (The pKa of acetic acid is 4.7.)
(A)
(B)
(C)
(D)
Acetate Species
pH
HC2H3O2
HC2H3O2
< 4.7
> 4.7
< 4.7
C2H3O2-
C2H3O2-
Name
First
801
Second
2,430
Third
3,660
Fourth
25,000
Fifth
32,820
12. The first five ionization energies of a
second-period element are listed in the table
above. Which of the following correctly
identifies the element and best explains the
data in the table?
(A)
(B)
(C)
(D)
> 4.7
B, because it has five core electrons
B, because it has three valence electrons
N, because it has five valence electrons
N, because it has three electrons in the
p sublevel
Molecular Molar Mass
Formula
(g/mol)
Ethane
C2H6
30
Butane
C4H10
58
Æ Mg2+(aq) + 2 OH-(aq)
Mg(OH)2(s) ¨
13. The exothermic dissolution of Mg(OH)2(s) in
water is represented by the equation above. The
Ksp of Mg(OH)2 is 1.8 ¥ 10-11. Which of the
following changes will increase the solubility of
Mg(OH)2 in an aqueous solution?
11. The molecular formula and molar mass of two
straight-chain hydrocarbons are listed in the table
above. Based on the information in the table,
which compound has the higher boiling point,
and why is that compound’s boiling point higher?
(A)
(B)
(C)
(D)
(A) C4H10 , because it has more hydrogen atoms,
resulting in more hydrogen bonding
(B) C4H10 , because it has more electrons,
resulting in greater polarizability and
stronger dispersion forces
(C) C2H6 , because its molecules are smaller and
they can get closer to one another, resulting
in stronger dispersion forces
(D) C2H6 , because its molecules are more polar,
resulting in stronger dipole-dipole
attractions
Unauthorized copying or reuse of
any part of this page is illegal.
Decreasing the pH
Increasing the pH
Adding NH3 to the solution
Adding Mg(NO3)2 to the solution
GO ON TO THE NEXT PAGE.
-10-
14. The heating curve for a sample of pure ethanol is provided
above. The temperature was recorded as a 50.0 g sample of
solid ethanol was heated at a constant rate. Which of the
following explains why the slope of segment T is greater
than the slope of segment R ?
(A) The specific heat capacity of the gaseous ethanol is less
than the specific heat capacity of liquid ethanol.
(B) The specific heat capacity of the gaseous ethanol is
greater than the specific heat capacity of liquid ethanol.
(C) The heat of vaporization of ethanol is less than the heat
of fusion of ethanol.
(D) The heat of vaporization of ethanol is greater than the
heat of fusion of ethanol.
H3PO4 Æ̈ H+ + H2PO4−
K a1 = 7.2 × 10−3
H2PO4− Æ̈ H+ + HPO42−
K a2 = 6.3 × 10−8
HPO42− Æ̈ H+ + PO43−
K a3 = 4.5 × 10−13
15. A solution is prepared by mixing 50 mL of 1 M NaH2PO4
with 50 mL of 1 M Na2HPO4 . On the basis of the
information above, which of the following species is present
in the solution at the lowest concentration?
(A) Na+
(B) HPO42−
(C) H2PO4
−
(D) PO43−
Unauthorized copying or reuse of
any part of this page is illegal.
GO ON TO THE NEXT PAGE. -11-
Æ X (g) + Y (g)
2 XY(g) ¨
2
2
Kp = 230
16. A certain gas, XY(g), decomposes as represented by the
equation above. A sample of each of the three gases is put
in a previously evacuated container. The initial partial
pressures of the gases are shown in the table below.
Gas
Initial Partial
Pressure (atm)
XY
0.010
X2
0.20
Y2
2.0
The temperature of the reaction mixture is held constant. In
which direction will the reaction proceed?
(A)
(B)
(C)
(D)
The reaction will form more products.
The reaction will form more reactant.
The mixture is at equilibrium, so there will be no change.
It cannot be determined unless the volume of the
container is known.
Questions 17-20 refer to the following information.
17. Which of the following changes will most likely
increase the rate of reaction between Li(s) and
water?
A 0.35 g sample of Li(s) is placed in an
Erlenmeyer flask containing 100 mL of water
at 25°C. A balloon is placed over the mouth of the
flask to collect the hydrogen gas that is generated.
(A) Using 125 mL of water instead of 100 mL
(B) Using a 0.25 g sample of Li(s) instead of a
0.35 g sample
(C) Using a 0.35 g sample of Li(s) cut into small
pieces
(D) Decreasing the water temperature before
adding the Li(s)
After all of the Li(s) has reacted with H2O(l), the
solution in the flask is added to a clean, dry buret and
used to titrate an aqueous solution of a monoprotic
acid. The pH curve for this titration is shown in the
diagram below.
18. What will be the effect on the amount of gas
produced if the experiment is repeated using
0.35 g of K(s) instead of 0.35 g of Li(s) ?
(A) No gas will be produced when K(s) is used.
(B) Some gas will be produced but less than the
amount of gas produced with Li(s).
(C) Equal quantities of gas will be produced with
the two metals.
(D) More gas will be produced with K(s) than
with Li(s).
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19. On the basis of the pH curve, the pKa value of the
acid is closest to
(A)
(B)
(C)
(D)
4
5
8
12
20. Which of the following is the balanced net-ionic equation for the reaction between Li(s) and water?
(A) 2 Li(s) + 2 H+(aq) + 2 OH−(aq) → 2 Li+(aq) + 2 OH−(aq) + H2(g)
(B) 2 Li(s) + 2 H2O(l) → 2 Li+(aq) + 2 OH−(aq) + H2(g)
(C) 2 Li(s) + 2 H2O(l) → 2 LiOH(s) + H2(g)
(D) 2 Li(s) + 2 H2O(l) → 2 LiH(s) + H2(g)
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-13-
23. A 1.0 g sample of a cashew was burned in a
calorimeter containing 1000. g of water, and the
temperature of the water changed from 20.0°C
to 25.0°C. In another experiment, a 3.0 g sample
of a marshmallow was burned in a calorimeter
containing 2000. g of water, and the temperature
of the water changed from 25.0°C to 30.0°C.
Based on the data, which of the following can be
concluded about the energy content for 1.0 g of
each of the two substances? (The specific heat of
water is 4.2 J/(g⋅°C).)
21. Benzene, C6H6 , has the structure shown above.
Considering the observation that benzene is only
sparingly soluble in water, which of the following
best describes the intermolecular forces of
attraction between water and benzene?
(A) The combustion of 1.0 g of cashew releases
less energy than the combustion of 1.0 g of
marshmallow.
(B) The combustion of 1.0 g of cashew releases
the same amount of energy as the
combustion of 1.0 g of marshmallow.
(C) The combustion of 1.0 g of cashew releases
more energy than the combustion of 1.0 g of
marshmallow.
(D) No comparison can be made because the
two systems started with different masses
of food, different masses of water, and
different initial temperatures.
(A) Benzene is nonpolar, therefore there are no
forces between water and benzene.
(B) The H atoms in benzene form hydrogen
bonds with the O atoms in water.
(C) Benzene is hydrophobic, therefore there is a
net repulsion between water and benzene.
(D) There are dipole-induced dipole and London
dispersion interactions between water and
benzene.
Solution
Acid
Ka
1
CH3CO2H
1.75 ¥ 10-5
2
CF3CO2H
1.0 ¥ 100
22. Acid-dissociation constants of two acids are
listed in the table above. A 20. mL sample of
a 0.10 M solution of each acid is titrated to the
equivalence point with 20. mL of 0.10 M NaOH.
Which of the following is a true statement about
the pH of the solutions at the equivalence point?
(A) Solution 1 has a higher pH at the equivalence
point because CH3CO2H is the stronger
acid.
(B) Solution 1 has a higher pH at the equivalence
point because CH3CO2H has the stronger
conjugate base.
(C) Solution 1 has a lower pH at the equivalence
point because CH3CO2H is the stronger
acid.
(D) Solution 1 has a lower pH at the equivalence
point because CH3CO2H has the stronger
conjugate base.
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24. The reaction between NO(g) and O2(g) to produce NO2(g) in a rigid reaction vessel is represented in the
diagram above. The pressure inside the container is recorded using a pressure gauge. Which of the following
statements correctly predicts the change in pressure as the reaction goes to completion at constant temperature,
and provides the correct explanation?
(A) The pressure will increase because the product molecules have a greater mass than either of the reactant
molecules.
(B) The pressure will decrease because there are fewer molecules of product than of reactants.
(C) The pressure will decrease because the product molecules have a lower average speed than the reactant
molecules.
(D) The pressure will not change because the total mass of the product molecules is the same as the total mass of
the reactant molecules.
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Questions 25-29 refer to the following information.
2 H2O2(aq) Æ 2 H2O(l) + O2(g)
DH° = −196 kJ/molrxn
The decomposition of H2O2(aq) is represented by the equation above. A student monitored the decomposition of a
1.0 L sample of H2O2(aq) at a constant temperature of 300. K and recorded the concentration of H2O2 as a function
of time. The results are given in the table below.
Time (s)
[H2O2]
0
2.7
200.
2.1
400.
1.7
600.
1.4
27. Which of the following statements is a correct
interpretation of the data regarding how the order
of the reaction can be determined?
25. Which of the following identifies the element(s)
being oxidized and reduced in the reaction?
(A)
(B)
(C)
(D)
Hydrogen is oxidized and oxygen is reduced.
Oxygen is oxidized and hydrogen is reduced.
Oxygen is both oxidized and reduced.
No elements are oxidized or reduced; the
reaction is not a redox reaction.
(A) The reaction must be first order because there
is only one reactant species.
(B) The reaction is first order if the plot of
ln [H2O2] versus time is a straight line.
(C) The reaction is first order if the plot of
1/[H2O2] versus time is a straight line.
26. The O2(g) produced from the decomposition of
the 1.0 L sample of H2O2(aq) is collected in a
previously evacuated 10.0 L flask at 300. K.
What is the approximate pressure in the flask
after 400. s? (For estimation purposes, assume
that 1.0 mole of gas in 1.0 L exerts a pressure
of 24 atm at 300. K.)
(D) The reaction is second order because 2 is
the coefficient of H2O2 in the chemical
equation.
28. The reaction is thermodynamically favorable. The
signs of ΔG° and ΔS° for the reaction are which
of the following?
(A) 1.2 atm
(B) 2.4 atm
(C) 12 atm
(D) 24 atm
ΔG°
(A) Positive
(B) Negative
(C) Positive
(D) Negative
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ΔS°
Positive
Positive
Negative
Negative
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32. A sample of a compound that contains only
the elements C, H, and N is completely burned
in O2 to produce 44.0 g of CO2 , 45.0 g of H2O,
and some NO2 . A possible empirical formula of
the compound is
29. Assume that the bond enthalpies of the oxygenhydrogen bonds in H2O are not significantly
different from those in H2O2 . Based on the value
of ΔH° of the reaction, which of the following
could be the bond enthalpies (in kJ/mol) for the
bonds broken and formed in the reaction?
(A)
(B)
(C)
(D)
O–O
in H2O2
300
150
500
250
O=O
in O2
500
500
300
300
(A)
(B)
(C)
(D)
O–H
500
500
150
150
CH2N
CH5N
C2H5N
C3H3N2
Æ H O+(aq) + F −(aq)
HF(aq) + H2O(l) ¨
3
33. The dissociation of the weak acid HF in water
is represented by the equation above. Adding a
1.0 mL sample of which of the following would
increase the percent ionization of HF(aq) in
10 mL of a solution of 1.0 M HF ?
30. Which of the following accounts for the
observation that the pH of pure water at 37°C
is 6.8 ?
(A)
(B)
(C)
(D)
(A) At 37∞C water is naturally acidic.
(B) At 37∞C the autoionization constant for
water, Kw , is larger than it is at 25∞C.
(C) At 37∞C water has a lower density than it
does at 25∞C; therefore, [H+] is greater.
(D) At 37∞C water ionizes to a lesser extent than
it does at 25∞C.
1.0 M KF
1.0 M H2SO4
10.0 M HF
Distilled water
31. To gravimetrically analyze the silver content of
a piece of jewelry made from an alloy of Ag and
Cu, a student dissolves a small preweighed
sample in HNO3(aq). Ag+(aq) and Cu2+(aq) ions
form in the solution. Which of the following
should be the next step in the analytical process?
(A) Centrifuging the solution to isolate the
heavier ions
(B) Evaporating the solution to recover the
dissolved nitrates
(C) Adding enough base solution to bring the pH
up to 7.0
(D) Adding a solution containing an anion that
forms an insoluble salt with only one of the
metal ions
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Questions 34-36 refer to the reactions represented below, which are involved in a demonstration commonly
known as “underwater fireworks.”
Reaction 1: CaC2(s) + 2 H2O(l)  C2H2(g) + Ca(OH)2(s)
Reaction 2: NaOCl(aq) + 2 HCl(aq)  Cl2(g) + NaCl(aq) + H2O(l)
Reaction 3: C2H2(g) + Cl2(g)  C2H2Cl2(g)
36. When Reaction 3 occurs, does the hybridization
of the carbon atoms change?
34. Ca(OH)2 (s) precipitates when a 1.0 g sample of
CaC2 (s) is added to 1.0 L of distilled water at
room temperature. If a 0.064 g sample of
CaC2 (s) (molar mass 64 g/mol) is used instead
and all of it reacts, which of the following will occur and why? (The value of K sp for Ca(OH)2
(A) Yes; it changes from sp to sp2 . (B) Yes; it changes from sp to sp3
. (C) Yes; it changes from sp2 to sp3 .
(D) No; it does not change.
is 8.0  108.)
(A) Ca(OH)2 will precipitate because Q > K sp .
(B) Ca(OH)2 will precipitate because Q < K sp .
(C) Ca(OH)2 will not precipitate because
Q > K sp .
(D) Ca(OH)2 will not precipitate because
Q < K sp .
35. Reaction 2 occurs when an excess of 6 M HCl(aq)
solution is added to 100. mL of NaOCl(aq) of
unknown concentration. If the reaction goes to
completion and 0.010 mol of Cl2(g) is produced,
then what was the molarity of the NaOCl(aq)
solution?
(A)
(B)
(C)
(D)
0.0010 M
0.010 M
0.10 M
1.0 M
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-18-
39. If equal masses of the following compounds
undergo complete combustion, which will yield
the greatest mass of CO2 ?
Ka
HC3H5O3(aq)
8.3 × 10− 4
CH3NH3+(aq)
2.3 × 10−11
(A)
(B)
(C)
(D)
37. The acid-dissociation constants of HC3H5O3(aq)
and CH3NH3+(aq) are given in the table above.
Which of the following mixtures is a buffer with a
pH of approximately 3 ?
Benzene, C6H6
Cyclohexane, C6H12
Glucose, C6H12O6
Methane, CH4
CaCO3(s) Æ̈
CaO(s) + CO2(g)
DH° = 178 kJ/molrxn
(A) A mixture of 100. mL of 0.1 M CH3NH3Cl
and 50. mL of 0.1 M NaOH
(B) A mixture of 100. mL of 0.1 M HC3H5O3
and 50. mL of 0.1 M NaOH
(C) A mixture of 100. mL of 0.1 M NaC3H5O3
and 100. mL of 0.1 M NaOH
(D) A mixture of 100. mL of 0.1 M CH3NH3Cl
and 100. mL of 0.1 M CH3NH2
40. The reaction system represented above is at
equilibrium. Which of the following will decrease
the amount of CaO(s) in the system?
(A) Increasing the volume of the reaction vessel
at constant temperature
(B) Lowering the temperature of the system
(C) Removing some CO2(g) at constant temperature (D) Removing some CaCO3(s) at constant
temperature
38. The Lewis electron-dot diagrams of the HClO3
molecule and the HClO2 molecule are shown
above at the left and right, respectively. Which of
the following statements identifies the stronger
acid and correctly identifies a factor that
contributes to its being the stronger acid?
(A) HClO3(aq) is the stronger acid because its
molecules experience stronger London
dispersion forces.
(B) HClO3(aq) is the stronger acid because the
additional electronegative oxygen atom on
the chlorine atom stabilizes the conjugate
base.
(C) HClO2(aq) is the stronger acid because its
molecules experience weaker London
dispersion forces.
(D) HClO2(aq) is the stronger acid because the
lone pairs of electrons on the chlorine atom
stabilize the conjugate base.
Acid
Solution
Volume of
NaOH Added
(mL)
A
40
B
75
C
115
D
200
41. To maximize the yield in a certain manufacturing
process, a solution of a weak monoprotic acid that
has a concentration between 0.20 M and 0.30 M
is required. Four 100. mL samples of the acid at
different concentrations are each titrated with a
0.20 M NaOH solution. The volume of NaOH
needed to reach the end point for each sample is
given in the table above. Which solution is the
most suitable to maximize the yield?
(A)
(B)
(C)
(D)
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Solution
Solution
Solution
Solution
A
B
C
D
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Questions 42-44 refer to the following information.
When free Cl(g) atoms encounter O3(g) molecules in the upper atmosphere, the following
reaction mechanism is proposed to occur.
Cl(g) + O3(g) → ClO(g) + O2(g)
slow step
ClO(g) + O3(g) → Cl(g) + 2 O2(g)
fast step
2 O3(g) → 3 O2(g)
overall reaction
ΔH = −285 kJ/molrxn
42. Which of the following rate laws for the overall
reaction corresponds to the proposed mechanism?
(A) Rate = k[O3]2
(B) Rate = k[Cl][O3]
(C) Rate = k[ClO][O3]2
(D) Rate = k
[O2 ]3
[O3 ]2
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X + O3 → XO + O2
XO + O3 → X + 2 O2
43. Which of the following reaction energy profiles
best corresponds to the proposed mechanism?
(A)
2 O3 → 3 O2
44. The proposed mechanism can be written in a more
general form, as shown above. Species other than
Cl can also decompose O3 through the same
mechanism. Which of the following chemical
species is most likely to decompose O3 in the
upper atmosphere through the above mechanism?
(A)
(B)
(C)
(D)
(B) He
Br
N2
O2
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-21-
HX(aq) + Y−(aq) Æ̈
HY(aq) + X−(aq)
Keq > 1
45. A solution of a salt of a weak acid HY is added to a solution
of another weak acid HX. Based on the information given
above, which of the following species is the strongest base?
(A)
(B)
(C)
(D)
HX(aq)
Y−(aq)
HY(aq)
X−(aq)
X(g) + 2 Q(g) Æ̈
R(g) + Z(g)
Kc = 1.3 × 105 at 50°C
46. A 1.0 mol sample of X(g) and a 1.0 mol sample of Q(g) are
introduced into an evacuated, rigid 10.0 L container and
allowed to reach equilibrium at 50°C according to the
equation above. At equilibrium, which of the following is true
about the concentrations of the gases?
1
[Q]
2
1
[X]
(B) [Q] =
2
(A) [R] =
(C) [R] = [Z] > [Q]
(D) [X] = [Q] = [R] = [Z]
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47. The potential energy of a system of two atoms as a function
of their internuclear distance is shown in the diagram above.
Which of the following is true regarding the forces between
the atoms when their internuclear distance is x ?
(A) The attractive and repulsive forces are balanced, so the
atoms will maintain an average internuclear distance x.
(B) There is a net repulsive force pushing the atoms apart, so
the atoms will move further apart.
(C) There is a net attractive force pulling the atoms together,
so the atoms will move closer together.
(D) It cannot be determined whether the forces between
atoms are balanced, attractive, or repulsive, because the
diagram shows only the potential energy.
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-23-
X(g) + 2 Y(g) Æ XY2(g)
48. In order to determine the order of the reaction represented
above, the initial rate of formation of XY2 is measured using
different initial values of [X] and [Y]. The results of the
experiment are shown in the table below.
Initial Rate of Formation of XY2
Trial
[X]
[Y]
1
0.50
0.50
8.0 ¥ 10-3
2
1.00
0.50
3.2 ¥ 10-2
3
1.00
1.00
6.4 ¥ 10-2
(M s-1)
In trial 2 which of the reactants would be consumed more
rapidly, and why?
(A)
(B)
(C)
(D)
X, because it has a higher molar concentration.
X, because the reaction is second order with respect to X.
Y, because the reaction is second order with respect to Y.
Y, because the rate of disappearance will be double
that of X.
49. In a paper chromatography experiment, a sample of a
pigment is separated into two components, X and Y, as
shown in the figure above. The surface of the paper is
moderately polar. What can be concluded about X and Y
based on the experimental results?
(A)
(B)
(C)
(D)
X
Y
X
Y
has a larger molar mass than Y does.
has a larger molar mass than X does.
is more polar than Y.
is more polar than X.
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-24-
50. The diagram above represents the absorption spectrum for a pure molecular substance. Which of the following
correctly indicates the type of transition observed for the substance in each of the regions of the absorption
spectrum?
(A)
(B)
(C)
(D)
Region X
Molecular vibration
Electronic transition
Molecular rotation
Electronic transition
Region Y
Molecular rotation
Molecular rotation
Molecular vibration
Molecular vibration
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any part of this page is illegal.
Region Z
Electronic transition
Molecular vibration
Electronic transition
Molecular rotation
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-25-
END OF SECTION I
IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS SECTION.
DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.
________________________________
MAKE SURE YOU HAVE DONE THE FOLLOWING.
• PLACED YOUR AP NUMBER LABEL ON YOUR ANSWER SHEET
• WRITTEN AND GRIDDED YOUR AP NUMBER CORRECTLY ON YOUR
ANSWER SHEET
• TAKEN THE AP EXAM LABEL FROM THE FRONT OF THIS BOOKLET
AND PLACED IT ON YOUR ANSWER SHEET
Unauthorized copying or reuse of
any part of this page is illegal.
-26-
Section II: Free-Response Questions
This is the free-response section of the 2015 AP exam.
It includes cover material and other administrative instructions
to help familiarize students with the mechanics of the exam.
(Note that future exams may differ in look from the following content.)
®
AP Chemistry Exam
2015
SECTION II: Free Response
DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.
At a Glance
Total Time
1 hour, 45 minutes
Number of Questions
7
Percent of Total Score
50%
Writing Instrument
Either pencil or pen with
black or dark blue ink
Electronic Device
Calculator allowed
Suggested Time
Approximately
23 minutes each for
questions 1 3 and
9 minutes each for
questions 4 7
Weight
Approximate weights:
Questions 1 3:
22% each
Questions 4 7:
9% each
Instructions
The questions for Section II are printed in this booklet. Pages containing a periodic table
and lists containing equations and constants are also printed in this booklet.
You may use the pages that the questions are printed on to organize your answers or for
scratch work, but you must write your answers in the areas designated for each response.
Only material written in the space provided will be scored.
Examples and equations may be included in your responses where appropriate. For
calculations, clearly show the method used and the steps involved in arriving at your
answers. You must show your work to receive credit for your answer. Pay attention to
significant figures.
Write clearly and legibly. Cross out any errors you make; erased or crossed-out work will
not be scored.
Manage your time carefully. You may proceed freely from one question to the next. You
may review your responses if you finish before the end of the exam is announced.
Form I
Form Code 4LBP4-S
25
-2-
®
AP CHEMISTRY EQUATIONS AND CONSTANTS
Throughout the exam the following symbols have the definitions specified unless otherwise noted.
L, mL
g
nm
atm
=
=
=
=
liter(s), milliliter(s)
gram(s)
nanometer(s)
atmosphere(s)
mm Hg
J, kJ
V
mol
=
=
=
=
ATOMIC STRUCTURE
millimeters of mercury
joule(s), kilojoule(s)
volt(s)
mole(s)
E = energy
 = frequency
 = wavelength
E = h
c = 
Planck’s constant, h = 6.626  1034 J s
Speed of light, c = 2.998  108 m s1
Avogadro’s number = 6.022  1023 mol1
Electron charge, e = 1.602 × 1019 coulomb
EQUILIBRIUM
Kc =
Kp =
[C]c [D]d
, where a A + b B  c C + d D
[A]a [B]b
Equilibrium Constants
Kc
Kp
Ka
Kb
Kw
(PC )c (PD )d
(PA )a (PB )b
+
Ka = [H ][A ]
[HA]
+
Kb = [OH ][HB ]
(molar concentrations)
(gas pressures)
(weak acid)
(weak base)
(water)
[B]
Kw = [H ][OH] = 1.0  1014 at 25C
= Ka  K b
+
pH = log[H+] , pOH = log[OH]
14 = pH + pOH
pH = pKa + log [A ]
[HA]
pKa = logKa , pKb = logKb
KINETICS
ln[A] t  ln[A] 0 =  kt
1 - 1
[A ]t
[A ]0
k = rate constant
t = time
t ½ = half-life
= kt
t ½ = 0.693
k
-3-
GASES, LIQUIDS, AND SOLUTIONS
PV = nRT
PA = Ptotal  XA, where XA =
moles A
total moles
Ptotal
= PA + PB + PC + . . .
n = m
M
K = C + 273
D= m
V
KE per molecule = 1 mv 2
P
V
T
n
m
M
D
KE
v
A
a
b
c
=
=
=
=
=
=
=
=
=
=
=
=
=
pressure
volume
temperature
number of moles
mass
molar mass
density
kinetic energy
velocity
absorbance
molar absorptivity
path length
concentration
2
Molarity, M = moles of solute per liter of solution
Gas constant, R = 8.314 J mol1 K1
= 0.08206 L atm mol1 K1
A = abc
= 62.36 L torr mol1 K1
1 atm = 760 mm Hg = 760 torr
STP = 273.15 K and 1.0 atm
Ideal gas at STP = 22.4 L mol−1
THERMODYNAMICS / ELECTROCHEMISTRY
q =
m=
c =
T=
S =
H =
G =
n =
E =
I =
q =
t =
q = mcT
 S products -  S reactants
H = Â DHf products - Â DHf reactants
S =
G =
 DGf products -  DGf reactants
G = H  TS
= RT ln K
=  nF E
I =
q
t
heat
mass
specific heat capacity
temperature
standard entropy
standard enthalpy
standard Gibbs free energy
number of moles
standard reduction potential
current (amperes)
charge (coulombs)
time (seconds)
Faraday’s constant, F = 96,485 coulombs per mole
of electrons
1 joule
1 volt =
1 coulomb
-4-
SECTION II BEGINS ON PAGE 6.
-5-
CHEMISTRY Section II 7 Questions Time—1 hour and 45 minutes YOU MAY USE YOUR CALCULATOR FOR THIS SECTION.
Directions: Questions 1–3 are long free-response questions that require about 23 minutes each to answer and are
worth 10 points each. Questions 4–7 are short free-response questions that require about 9 minutes each to answer
and are worth 4 points each.
Write your response in the space provided following each question. Examples and equations may be included in
your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving
at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures.
+
Half-Reaction
E° (V)
−
− 0.085
2 CO2(g) + 12 H (aq) + 12 e
Æ C2H5OH(aq) + 3 H2O(l)
O2(g) + 4 H+(aq) + 4 e− Æ 2 H2O(l)
1.229
1. A student uses a galvanic cell to determine the concentration of ethanol, C2H5OH, in an aqueous solution. The
cell is based on the half-cell reactions represented in the table above.
(a) Write a balanced equation for the overall reaction that occurs in the cell.
(b) Calculate E° for the overall reaction that occurs in the cell.
(c) A 10.0 mL sample of C2H5OH(aq) is put into the electrochemical cell. The cell produces an average current
of 0.10 amp for 20. seconds, at which point the C2H5OH(aq) has been totally consumed.
(i) Calculate the charge, in coulombs, that passed through the cell.
(ii) Calculate the initial [C2H5OH] in the solution.
An alternative approach to determine the concentration of C2H5OH(aq) in a solution is based on the reaction
represented below.
3 C2H5OH(aq) + 2 Cr2O72−(aq) + 16 H+(aq) Æ 4 Cr3+(aq) + 3 CH3COOH(aq) + 11 H2O(l)
orange
blue-green
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A solution has an initial Cr2O72−(aq) concentration of 1.0 × 10−3 M and an initial C2H5OH(aq) concentration
of 0.500 M. The solution contains enough strong acid to keep the pH essentially constant throughout the
reaction. The student places a sample of the solution in a cuvette that has a path length of 0.50 cm and places it
in a spectrophotometer set to measure absorbance at 440 nm. (Cr2O72−(aq) is the only species in the reaction
mixture that absorbs light at this wavelength.) The absorbance of Cr2O72−(aq) in the solution is monitored as the
reaction proceeds; the table below shows the absorbance as a function of time for the first trial.
Time (min)
Absorbance at 440 nm
0.00
0.782
1.50
0.553
3.00
0.389
4.50
0.278
6.00
0.194
(d) Calculate the value of [Cr2O72−] at 1.50 min.
(e) The student runs a second trial but this time uses a cuvette that has a path length of 1.00 cm. Describe how
the experimental setup should be adjusted to keep the initial absorbance at 0.782. Justify your answer with
respect to the factors that influence the absorbance of a sample in a spectrophotometer.
For the concentrations of reactants used in the experiment, the rate of the reaction can be written as follows.
Rate = kobserved[Cr2O72−]y, where kobserved = k[C2H5OH]
(f) Explain how the experimental data indicate that the reaction is first order with respect to Cr2O72−.
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-9-
2. Answer the following questions about Fe and Al compounds.
(a) Fe2O3(s) and Al2O3(s) have similar chemical properties; some similarities are due to the oxides having
similar lattice energies. Give two reasons why the lattice energies of the oxides are similar.
Use the following reactions that involve Fe and Al compounds to answer parts (b) and (c).
In distilled water
Reaction 1:
Fe2O3(s) + 3 H2O(l) Æ 2 Fe(OH)3(s)
Reaction 2:
Al2O3(s) + 3 H2O(l) Æ 2 Al(OH)3(s)
In base
Reaction 3:
Fe(OH)3(s) + NaOH(aq) Æ no reaction
Reaction 4:
Al(OH)3(s) + NaOH(aq) Æ NaAl(OH)4(aq)
In acid
Reaction 5:
Fe(OH)3(s) + 3 HCl(aq) Æ FeCl3(aq) + 3 H2O(l)
Reaction 6:
Al(OH)3(s) + 3 HCl(aq) Æ AlCl3(aq) + 3 H2O(l)
Reaction 7:
NaAl(OH)4(aq) + HCl(aq) Æ Al(OH)3(s) + NaCl(aq) + H2O(l)
When heated
Reaction 8:
heat
Æ Fe2O3(s) + 3 H2O(g)
2 Fe(OH)3(s) æææ
Reaction 9:
heat
Æ Al2O3(s) + 3 H2O(g)
2 Al(OH)3(s) æææ
Compound
Ksp
Fe(OH)3
4 × 10−38
Al(OH)3
1 × 10−33
(b) The Ksp values for Fe(OH)3 and Al(OH)3 are given in the table above. A 1.0 g sample of powdered
Fe2O3(s) and a 1.0 g sample of powdered Al2O3(s) are mixed together and placed in 1.0 L of distilled water.
(i) Which ion, Fe3+(aq) or Al3+(aq), will be present in the higher concentration? Justify your answer with
respect to the Ksp values provided.
(ii) Write a balanced chemical equation for the dissolution reaction that results in the production of the ion
that you identified in part (i).
(c) Students are asked to develop a plan for separating Al2O3(s) from a mixture of powdered Fe2O3(s) and
powdered Al2O3(s) using chemical reactions and laboratory techniques.
(i) One student proposes that Al2O3(s) can be separated from the mixture by adding water to the mixture
and then filtering. Explain why this approach is not reasonable.
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(ii) A second student organizes a plan using a table. The first two steps have already been entered in the
table, as shown below. Complete the plan by listing the additional steps that are needed to recover the
Al2O3(s). List the steps in the correct order and refer to the appropriate reaction by number, if
applicable.
Step
Description
Reaction(s)
1
Add NaOH(aq) to convert Al2O3(s) to Al(OH)3(s) and then to NaAl(OH)4(aq).
2 and 4
2
Filter out the solid Fe(OH)3 from the mixture and save the filtrate.
−
3
4
5
(iii) The second student recovers 5.5 g of Al2O3(s) from a 10.0 g sample of the mixture. Calculate the
percent of Al by mass in the mixture of the two powdered oxides. (The molar mass of Al2O3 is
101.96 g/mol, and the molar mass of Fe2O3 is 159.70 g/mol.)
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ADDITIONAL PAGE FOR ANSWERING QUESTION 2 GO ON TO THE NEXT PAGE.
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ADDITIONAL PAGE FOR ANSWERING QUESTION 2 GO ON TO THE NEXT PAGE.
-13-
3. The structures of two compounds commonly found in food, lauric acid, C12H24O2 , and sucrose, C12H22O11 , are
shown above.
(a) Which compound, lauric acid or sucrose, is more soluble in water? Justify your answer in terms of the
intermolecular forces present between water and each of the compounds.
(b) Assume that a 1.5 g sample of lauric acid is combusted and all of the heat energy released is transferred to a
325 g sample of water initially at 25°C. Calculate the final temperature of the water if DHcombustion of lauric
acid is −37 kJ/g and the specific heat of water is 4.18 J/(g⋅K).
(c) In an attempt to determine DHcombustion of lauric acid experimentally, a student places a 1.5 g sample of
lauric acid in a ceramic dish underneath a can made of Al containing 325 g of water at 25°C. The student
ignites the sample of lauric acid with a match and records the highest temperature reached by the water in
the can.
(i) The experiment is repeated using a can of the same mass, but this time the can is made of Cu. The
specific heat of Cu is 0.39 J/(g⋅K), and the specific heat of Al is 0.90 J/(g⋅K). Will the final
temperature of the water in the can made of Cu be greater than, less than, or equal to the final
temperature of the water in the can made of Al? Justify your answer.
(ii) In both experiments it was observed that the measured final temperature of the water was less than the
final temperature calculated in part (b). Identify one source of experimental error that might account
for this discrepancy and explain why the error would make the measured final temperature of the water
lower than predicted.
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(d) The experiment described above is repeated using a 1.5 g sample of sucrose. The combustion reaction for
sucrose in air is represented below.
C12H22O11(s) + 12 O2(g) Æ 12 CO2(g) + 11 H2O(g)
(i) Even though DG° for the combustion of sucrose in air has a value of −5837 kJ/molrxn , the combustion
reaction does not take place unless it is ignited. Explain.
(ii) Predict the sign of DS° for the reaction and justify your prediction.
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ADDITIONAL PAGE FOR ANSWERING QUESTION 3 GO ON TO THE NEXT PAGE.
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ADDITIONAL PAGE FOR ANSWERING QUESTION 3 GO ON TO THE NEXT PAGE.
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Molecule
Boiling Point of Compound
(K)
Dipole Moment
(debyes)
Polarizability
(10−24 cm3)
HCl
188
1.05
2.63
HBr
207
0.80
3.61
HI
238
0.38
5.44
4. The boiling points, dipole moments, and polarizabilities of three hydrogen halides are given in the table above.
(a) Based on the data in the table, what type of intermolecular force among the molecules HCl(l), HBr(l), and
HI(l) is able to account for the trend in boiling points? Justify your answer.
(b) Based on the data in the table, a student predicts that the boiling point of HF should be 174 K. The observed
boiling point of HF is 293 K. Explain the failure of the student’s prediction in terms of the types and
strengths of the intermolecular forces that exist among HF molecules.
(c) A representation of five molecules of HBr in the liquid state is shown in box 1 below. In box 2, draw a
representation of the five molecules of HBr after complete vaporization has occurred.
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ADDITIONAL PAGE FOR ANSWERING QUESTION 4 GO ON TO THE NEXT PAGE.
-19-
5. Answer the following questions about two isomers, methyl methanoate and ethanoic acid. The molecular
formula of the compounds is C2H4O2 .
(a) Complete the Lewis electron-dot diagram of methyl methanoate in the box below. Show all valence
electrons.
A student puts 0.020 mol of methyl methanoate into a previously evacuated rigid 1.0 L vessel at 450 K. The
pressure is measured to be 0.74 atm. When the experiment is repeated using 0.020 mol of ethanoic acid instead
of methyl methanoate, the measured pressure is lower than 0.74 atm. The lower pressure for ethanoic acid is due
to the following reversible reaction.
CH3COOH(g) + CH3COOH(g ) R (CH3COOH)2 (g)
(b) Assume that when equilibrium has been reached, 50. percent of the ethanoic acid molecules have reacted.
(i) Calculate the total pressure in the vessel at equilibrium at 450 K.
(ii) Calculate the value of the equilibrium constant, Kp , for the reaction at 450 K.
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ADDITIONAL PAGE FOR ANSWERING QUESTION 5 GO ON TO THE NEXT PAGE.
-21-
Peak 1
Peak 2
6.72 ¥ 10 4 kJ/mol
Peak 3
3.88 ¥ 103 kJ/mol
1.68 ¥ 103 kJ/mol
6. The complete photoelectron spectrum of an unknown element is shown above. The frequency ranges of different
regions of the electromagnetic spectrum are given in the table below.
Region of Electromagnetic Spectrum
Frequency Range (s−1)
1 × 1012
Infrared (IR)
to 4 × 1014
Ultraviolet/visible (UV/vis)
4 × 1014 to 5 × 1016
X-rays
5 × 1016 to 1 × 1019
Gamma rays
> 1 × 1019
(a) To generate the spectrum above, a source capable of producing electromagnetic radiation with an energy
of 7 × 104 kJ per mole of photons was used. Such radiation is from which region of the electromagnetic
spectrum? Justify your answer with a calculation.
(b) A student examines the spectrum and proposes that the second ionization energy of the element is
3.88 × 103 kJ/mol. To refute the proposed interpretation of the spectrum, identify the following.
(i) The subshell from which an electron is removed in the second ionization of an atom of the element
(ii) The subshell that corresponds to the second peak of the photoelectron spectrum above
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-23-
HC9H 7O 4 (aq ) + H 2 O(l ) R H 3O + (aq ) + C9H 7O 4 - (aq )
7. The molecular formula of acetylsalicylic acid, also known as aspirin, is HC9H7O4 . The dissociation of
HC9H7O4(aq) is represented by the equation above. The pH of 0.0100 M HC9H7O4(aq) is measured to be 2.78.
(a) Write the expression for the equilibrium constant, Ka , for the reaction above.
(b) Calculate the value of Ka for acetylsalicylic acid.
(c) An aqueous solution of aspirin is buffered to have equal concentrations of HC9H7O4(aq) and C9H7O4−(aq).
Calculate the pH of the solution.
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ADDITIONAL PAGE FOR ANSWERING QUESTION 7 GO ON TO THE NEXT PAGE.
-25-
STOP
END OF EXAM
IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS SECTION. ________________________________
THE FOLLOWING INSTRUCTIONS APPLY TO THE COVERS OF THE
SECTION II BOOKLET.
• MAKE SURE YOU HAVE COMPLETED THE IDENTIFICATION
INFORMATION AS REQUESTED ON THE FRONT AND BACK
COVERS OF THE SECTION II BOOKLET.
•
CHECK TO SEE THAT YOUR AP NUMBER LABEL APPEARS
IN THE BOX ON THE FRONT COVER.
• MAKE SURE YOU HAVE USED THE SAME SET OF AP
NUMBER LABELS ON ALL AP EXAMS YOU HAVE TAKEN
THIS YEAR.
-26-
Multiple-Choice Answer Key
The following contains the answers to
the multiple-choice questions in this exam.
Answer Key for AP Chemistry
Practice Exam, Section I
Question 1: A
Question 26: A
Question 2: C
Question 27: B
Question 3: D
Question 28: B
Question 4: A
Question 29: B
Question 5: C
Question 30: B
Question 6: C
Question 31: D
Question 7: C
Question 32: B
Question 8: B
Question 33: D
Question 9: A
Question 34: D
Question 10: D
Question 35: C
Question 11: B
Question 36: A
Question 12: B
Question 37: B
Question 13: A
Question 38: B
Question 14: A
Question 39: A
Question 15: D
Question 40: B
Question 16: B
Question 41: C
Question 17: C
Question 42: B
Question 18: B
Question 43: B
Question 19: B
Question 44: B
Question 20: B
Question 45: B
Question 21: D
Question 46: C
Question 22: B
Question 47: A
Question 23: C
Question 48: D
Question 24: B
Question 49: D
Question 25: C
Question 50: D
Free-Response Scoring Guidelines
The following contains the scoring guidelines
for the free-response questions in this exam.
AP® CHEMISTRY
2015 SCORING GUIDELINES
Question 1
E ° (V)
Half-Reaction
2 CO2(g) + 12
H+(aq)
+ 12
e−
Æ C2H5OH(aq) + 3 H2O(l )
− 0.085
O2(g) + 4 H+(aq) + 4 e− Æ 2 H2O(l )
1.229
A student uses a galvanic cell to determine the concentration of ethanol, C2H5OH(aq), in an aqueous solution.
The cell is based on the half-cell reactions represented in the table above.
(a) Write a balanced equation for the overall reaction that occurs in the cell.
1 point is earned for the
correct reactants and products.
C2H5OH(aq) + 3 O2(g) → 2 CO2(g) + 3 H2O(l )
1 point is earned for balancing the equation.
(b) Calculate E ° for the overall reaction that occurs in the cell.
E ° = −(−0.085 V) + 1.229 V = +1.314 V
1 point is earned for a correct answer that
is consistent with the equation in part (a).
(c) A 10.0 mL sample of C2H5OH(aq) is put into the electrochemical cell. The cell produces an average
current of 0.10 amp for 20. seconds, at which point the C2H5OH(aq) has been totally consumed.
(i) Calculate the charge, in coulombs, that passed through the cell.
I=
q
⇒ q = I t = 0.10 amp × 20. s = 2.0 C
t
1 point is earned for the correct charge.
(ii) Calculate the initial [C2H5OH] in the solution.
2.0 C ×
1 mol C2 H 5OH
1 mol e ×
= 1.7 × 10−6 mol C2H5OH
96, 485 C
12 mol e -
1.7 × 10 -6 mol C2 H5OH 1000 mL
= 1.7 × 10−4 M
×
10.0 mL
1.0 L
1 point is earned for the
number of moles of C2H5OH.
1 point is earned for the
initial molarity of C2H5OH.
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Question 1 (continued)
An alternative approach to determine the concentration of C2H5OH(aq) in a solution is based on the reaction
represented below.
3 C2H5OH(aq) + 2 Cr2O72−(aq) + 16 H+(aq) Æ 4 Cr3+(aq) + 3 CH3COOH(aq) + 11 H2O(l )
orange
blue-green
A solution has an initial Cr2O72−(aq) concentration of 1.0 ×10−3 M and an initial C2H5OH(aq) concentration
of 0.500 M. The solution contains enough strong acid to keep the pH essentially constant throughout the
reaction. The student places a sample of the solution in a cuvette that has a path length of 0.50 cm and places
it in a spectrophotometer set to measure absorbance at 440 nm. (Cr2O72−(aq) is the only species in the reaction
mixture that absorbs light at this wavelength.) The absorbance of Cr2O72−(aq) in the solution is monitored as
the reaction proceeds; the table below shows the absorbance as a function of time for the first trial.
Time (min)
Absorbance at 440 nm
0.00
0.782
1.50
0.553
3.00
0.389
4.50
0.278
6.00
0.194
(d) Calculate the value of [Cr2O72−] at 1.50 min.
Absorbance is proportional to [Cr2O72−].
1.0 × 10-3 M
x
=
0.782
0.553
x = 7.1 × 10−4 M
1 point is earned for the
correct concentration.
OR
Initial condition: a =
At 1.50 min:
c =
A
0.782
=
= 1564 M -1 cm -1
bc (0.50 cm)(1.0 × 10-3 M )
A
0.553
=
= 7.1 × 10-4 M
ab (1564 M -1 cm -1 )(0.50 cm)
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2015 SCORING GUIDELINES
Question 1 (continued)
(e) The student runs a second trial but this time uses a cuvette that has a path length of 1.00 cm. Describe
how the experimental setup should be adjusted to keep the initial absorbance at 0.782. Justify your
answer with respect to the factors that influence the absorbance of a sample in a spectrophotometer.
A = abc
If path length (b) is doubled, and molar absorptivity (a) is
constant, the initial concentration of Cr2O72− (c) must be
halved to keep the initial absorbance (A) constant.
1 point is earned for reference
to a factor that affects absorbance.
1 point is earned for the correct adjustment.
For the concentrations of reactants used in the experiment, the rate of the reaction can be written as follows.
Rate = kobserved[Cr2O72−]y, where kobserved = k[C2H5OH]
(f) Explain how the experimental data indicate that the reaction is first order with respect to Cr2O72−.
Absorbance is proportional to concentration of Cr2O72−. Absorbance is
halved after 3.00 min and again after another 3.00 min. Thus the
half-life of the reaction is constant, so the reaction must be first order
with respect to Cr2O72−.
1 point is earned for a correct
explanation that uses the data.
OR
Demonstration that the rate of change in ln(A) over time is constant.
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Question 2
Answer the following questions about Fe and Al compounds.
(a) Fe2O3(s) and Al2O3(s) have similar chemical properties; some similarities are due to the oxides having
similar lattice energies. Give two reasons why the lattice energies of the oxides are similar.
Fe3+ and Al3+ have similar sizes (radii).
1 point is earned for each reason.
Fe3+ and Al3+ have the same charge.
Use the following reactions that involve Fe and Al compounds to answer parts (b) and (c).
In distilled water
Reaction 1:
Fe2O3(s) + 3 H2O(l ) Æ 2 Fe(OH)3(s)
Reaction 2:
Al2O3(s) + 3 H2O(l ) Æ 2 Al(OH)3(s)
In base
Reaction 3:
Fe(OH)3(s) + NaOH(aq) Æ no reaction
Reaction 4:
Al(OH)3(s) + NaOH(aq) Æ NaAl(OH)4(aq)
In acid
Reaction 5:
Fe(OH)3(s) + 3 HCl(aq) Æ FeCl3(aq) + 3 H2O(l )
Reaction 6:
Al(OH)3(s) + 3 HCl(aq) Æ AlCl3(aq) + 3 H2O(l )
Reaction 7:
NaAl(OH)4(aq) + HCl(aq) Æ Al(OH)3(s) + NaCl(aq) + H2O(l )
When heated
Reaction 8:
heat
2 Fe(OH)3(s) æææ
Æ Fe2O3(s) + 3 H2O(g)
Reaction 9:
heat
Æ Al2O3(s) + 3 H2O(g)
2 Al(OH)3(s) æææ
Compound
Ksp
Fe(OH)3
4 × 10−38
Al(OH)3
1 × 10−33
(b) The Ksp values for Fe(OH)3 and Al(OH)3 are given in the table above. A 1.0 g sample of powdered
Fe2O3(s) and a 1.0 g sample of powdered Al2O3(s) are mixed together and placed in 1.0 L of distilled
water.
(i) Which ion, Fe3+(aq) or Al3+(aq), will be present in the higher concentration? Justify your answer
with respect to the Ksp values provided.
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Question 2 (continued)
Al3+ will be present in higher concentration.
Al(OH)3 has the same stoichiometry as Fe(OH)3 but a
greater Ksp .
1 point is earned for the
correct choice and explanation.
(ii) Write a balanced chemical equation for the dissolution reaction that results in the production of the
ion that you identified in part (i).
Al(OH)3(s) → Al3+(aq) + 3 OH−(aq)
OR
1 point is earned for a balanced equation.
Al2O3(s) + 3 H2O(l) → 2 Al3+(aq) + 6 OH−(aq)
(c) Students are asked to develop a plan for separating Al2O3(s) from a mixture of powdered Fe2O3(s)
and powdered Al2O3(s) using chemical reactions and laboratory techniques.
(i) One student proposes that Al2O3(s) can be separated from the mixture by adding water to
the mixture and then filtering. Explain why this approach is not reasonable.
This approach only works when there is a
significant difference in water solubility
between two substances.
1 point is earned for a correct explanation.
(ii) A second student organizes a plan using a table. The first two steps have already been entered in
the table, as shown below. Complete the plan by listing the additional steps that are needed to
recover the Al2O3(s). List the steps in the correct order and refer to the appropriate reaction by
number, if applicable.
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Question 2 (continued)
Step
Description
Reaction(s)
1
Add NaOH(aq) to convert Al2O3(s) to Al(OH)3(s) and then to NaAl(OH)4(aq).
2 and 4
2
Filter out the solid Fe(OH)3 from the mixture and save the filtrate.
−
3
Add HCl to the filtrate until a precipitate of Al(OH)3 forms.
7
4
Filter out the solid Al(OH)3 .
−
5
Heat the solid Al(OH)3 to form Al2O3 .
9
(See rows 3-5 in table.)
1 point is earned for each correct row
(description plus reaction number, if applicable).
(iii) The second student recovers 5.5 g of Al2O3(s) from a 10.0 g sample of the mixture.
Calculate the percent of Al by mass in the mixture of the two powdered oxides.
(The molar mass of Al2O3 is 101.96 g/mol, and the molar mass of Fe2O3 is 159.70 g/mol.)
5.5 g Al2 O3 ×
1 mol Al2 O3
26.98 g Al
2 mol Al
×
×
= 2.9 g Al
101.96 g Al2 O3 1 mol Al2 O3
1 mol Al
2.9 g
× 100 = 29%
10.0 g
1 point is earned for the
number of grams of Al.
1 point is earned for
the mass percent Al.
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Question 3
The structures of two compounds commonly found in food, lauric acid, C12H24O2 , and sucrose, C12H22O11 ,
are shown above.
(a) Which compound, lauric acid or sucrose, is more soluble in water? Justify your answer in terms of the
intermolecular forces present between water and each of the compounds.
Sucrose is more soluble than lauric acid. Stronger
interactions occur between sucrose and water
molecules due to a greater capacity for hydrogen
bonding as a result of a larger number of −OH
groups in sucrose. Although lauric acid molecules
have one site for hydrogen bonding, the long
hydrocarbon chain causes London dispersion forces
to be the predominant, yet weaker, interaction with
water molecules.
1 point is earned for discussion of hydrogen
bonding between each substance and water.
1 point is earned for a valid conclusion about
solubility based on a comparison of these forces.
(b) Assume that a 1.5 g sample of lauric acid is combusted and all of the heat energy released is transferred to
a 325 g sample of water initially at 25°C. Calculate the final temperature of the water if ∆Hcombustion of
lauric acid is −37 kJ/g and the specific heat of water is 4.18 J/(g⋅K).
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Question 3 (continued)
1 point is earned
for heat released.
37 kJ/g × 1.5 g = 56 kJ released
q = mc∆T ⇒ ∆T =
q
56,000 J
=
= 41 K (41°C)
mc
(325 g)(4.18 J/(g◊ K))
1 point is earned
for a final temperature
consistent with the heat
released.
Final temperature = 25°C + 41°C = 66°C
(c) In an attempt to determine ∆Hcombustion of lauric acid experimentally, a student places a 1.5 g sample of
lauric acid in a ceramic dish underneath a can made of Al containing 325 g of water at 25°C. The student
ignites the sample of lauric acid with a match and records the highest temperature reached by the water in
the can.
(i) The experiment is repeated using a can of the same mass, but this time the can is made of Cu.
The specific heat of Cu is 0.39 J/(g⋅K), and the specific heat of Al is 0.90 J/(g⋅K). Will the final
temperature of the water in the can made of Cu be greater than, less than, or equal to the final
temperature of the water in the can made of Al? Justify your answer.
The temperature will be greater in the can made of
Cu. Since Cu has a lower specific heat than Al,
more heat will be absorbed by the water in the Cu
can than the water in the Al can.
1 point is earned for the correct answer.
1 point is earned for an appropriate explanation.
(ii) In both experiments it was observed that the measured final temperature of the water was less than
the final temperature calculated in part (b). Identify one source of experimental error that might
account for this discrepancy and explain why the error would make the measured final temperature
of the water lower than predicted.
Possible answers include:
Heat is lost to the environment. The temperature of the water is lower
because less heat is transferred to the water.
OR
1 point is earned for a valid
source of error.
Heat is transferred to the can or thermometer. The temperature of the
water is lower because less heat is transferred to the water.
1 point is earned for an
acceptable explanation.
OR
Incomplete combustion of the lauric acid. The temperature of the
water is lower because less heat was released from the combustion
reaction.
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AP® CHEMISTRY
2015 SCORING GUIDELINES
Question 3 (continued)
(d) The experiment described above is repeated using a 1.5 g sample of sucrose. The combustion reaction for
sucrose in air is represented below.
C12H22O11(s) + 12 O2(g) Æ 12 CO2(g) + 11 H2O(g)
(i) Even though ∆G ° for the combustion of sucrose in air has a value of −5837 kJ/molrxn , the
combustion reaction does not take place unless it is ignited. Explain.
The reaction has a high activation energy. The ignition
source provides energy to the molecules; some of them
can then overcome the activation energy barrier. The
reaction is exothermic, so the heat released allows the
reaction to continue.
1 point is earned for a correct explanation.
(ii) Predict the sign of ∆S ° for the reaction and justify your prediction.
∆S ° for the reaction is positive; there are 23 moles of
gaseous products for every 12 moles of gaseous reactant.
1 point is earned for the
correct sign and justification.
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AP® CHEMISTRY
2015 SCORING GUIDELINES
Question 4
Molecule
Boiling Point of Compound
(K)
Dipole Moment
(debyes)
Polarizability
(10−24 cm3)
HCl
188
1.05
2.63
HBr
207
0.80
3.61
HI
238
0.38
5.44
The boiling points, dipole moments, and polarizabilities of three hydrogen halides are given in
the table above.
(a) Based on the data in the table, what type of intermolecular force among the molecules HCl(l ), HBr(l ),
and HI(l ) is able to account for the trend in boiling points? Justify your answer.
London dispersion forces account for the trend in boiling
points.
As polarizability of the molecules increases, so does the
boiling point of the substance.
London dispersion forces depend upon the polarizability
of molecules, thus more polarizable molecules have
stronger intermolecular forces.
1 point is earned for identifying
London dispersion forces.
1 point is earned for explaining the link
between polarizability and boiling point.
(b) Based on the data in the table, a student predicts that the boiling point of HF should be 174 K.
The observed boiling point of HF is 293 K. Explain the failure of the student’s prediction in terms
of the types and strengths of the intermolecular forces that exist among HF molecules.
HF molecules have London dispersion forces and
hydrogen bonding. The hydrogen bonding is stronger
than both regular dipole-dipole forces and London
dispersion forces and accounts for HF’s higher boiling
point.
1 point is earned for a correct explanation.
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Question 4 (continued)
(c) A representation of five molecules of HBr in the liquid state is shown in box 1 below. In box 2,
draw a representation of the five molecules of HBr after complete vaporization has occurred.
The drawing in Box 2 should show the undissociated
molecules randomly distributed throughout the box.
1 point is earned for an acceptable
diagram.
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AP® CHEMISTRY
2015 SCORING GUIDELINES
Question 5
Answer the following questions about two isomers, methyl methanoate and ethanoic acid. The molecular
formula of the compounds is C2H4O2 .
(a) Complete the Lewis electron-dot diagram of methyl methanoate in the box below. Show all valence
electrons.
See diagram above.
(Note: A pair of electrons can be drawn as a pair of dots or a line segment.)
1 point is earned for
a correct diagram.
A student puts 0.020 mol of methyl methanoate into a previously evacuated rigid 1.0 L vessel at 450 K.
The pressure is measured to be 0.74 atm. When the experiment is repeated using 0.020 mol of ethanoic acid
instead of methyl methanoate, the measured pressure is lower than 0.74 atm. The lower pressure for ethanoic
acid is due to the following reversible reaction.
CH3COOH(g) + CH3COOH(g) (CH3COOH)2(g)
(b) Assume that when equilibrium has been reached, 50. percent of the ethanoic acid molecules have reacted.
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2015 SCORING GUIDELINES
Question 5 (continued)
(i) Calculate the total pressure in the vessel at equilibrium at 450 K.
Let ninitial be the number of moles of gas particles before any reaction occurs.
Because 50. percent of the molecules reacted, nfinal = 0.75 ninitial .
P final
Pinitial
=
n initial
n final
0.020 mol
= 0.010 mol
2
0.010 mol
n(CH3COOH)2 =
= 0.0050 mol
2
nCH3COOH =
Pfinal =
( Pinitial )( n final )
n initial
=
1 point is earned for
the correct pressure.
(0.74 atm)(0.015 mol gas particles)
= 0.56 atm
0.020 mol gas particles
OR
PV = nRT
P =
(0.015 mol)(0.08206 L atm mol -1 K -1 )(450 K)
= 0.55 atm
1.0 L
(ii) Calculate the value of the equilibrium constant, Kp , for the reaction at 450 K.
P = 0.56 atm
Mole fraction CH3COOH =
0.010 mol
2
=
0.015 mol
3
Mole fraction (CH3COOH)2 =
0.0050 mol
1
=
0.015 mol
3
2
(0.56 atm) = 0.37 atm
3
1
= (0.56 atm) = 0.19 atm
3
PCH3COOH =
P(CH3COOH)2
Kp =
( P(CH 3COOH)2 )
( P(CH 3COOH) )
2
=
1 point is earned for the
correct partial pressures.
1 point is earned for an answer
that uses partial pressures correctly.
0.19
= 1.4
(0.37)2
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2015 SCORING GUIDELINES
Question 6
Peak 1
Peak 2
4
3
6.72 × 10 kJ/mol
Peak 3
1.68 × 103 kJ/mol
3.88 × 10 kJ/mol
The complete photoelectron spectrum of an unknown element is shown above. The frequency ranges
of different regions of the electromagnetic spectrum are given in the table below.
Region of Electromagnetic Spectrum
Infrared (IR)
Frequency Range (s−1)
1 × 1012
to 4 × 1014
Ultraviolet/visible (UV/vis)
4 × 1014 to 5 × 1016
X-rays
5 × 1016 to 1 × 1019
Gamma rays
> 1 × 1019
(a) To generate the spectrum above, a source capable of producing electromagnetic radiation with an energy
of 7 × 104 kJ per mole of photons was used. Such radiation is from which region of the electromagnetic
spectrum? Justify your answer with a calculation.
1 mol photons
kJ
7 × 104 kJ
×
= 1.16 × 10-19
1 mol photons 6.022 × 1023 photons
photon
= 1 × 10-19
n =
J
kJ
1000 J
= 1 × 10-16
×
1 kJ
photon
photon
1.16 × 10-16 J
E
=
= 1.75 × 1017 s-1 = 2 × 1017 s-1
h
6.626 × 10-34 J s
1 point is earned for
calculating the amount of energy
per photon.
1 point is earned for the frequency
and region of the spectrum.
The radiation is in the X-ray region.
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2015 SCORING GUIDELINES
Question 6 (continued)
(b) A student examines the spectrum and proposes that the second ionization energy of the element is
3.88 × 103 kJ/mol. To refute the proposed interpretation of the spectrum, identify the following.
(i) The subshell from which an electron is removed in the second ionization of an atom of the element
2p
1 point is earned for the correct answer.
(ii) The subshell that corresponds to the second peak of the photoelectron spectrum above
2s
1 point is earned for the correct answer.
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2015 SCORING GUIDELINES
Question 7
HC9 H 7O4 (aq) + H 2O(l ) H 3O + (aq) + C9 H 7O4 - (aq)
The molecular formula of acetylsalicylic acid, also known as aspirin, is HC9H7O4 . The dissociation of
HC9H7O4(aq) is represented by the equation above. The pH of 0.0100 M HC9H7O4(aq) is measured to
be 2.78.
(a) Write the expression for the equilibrium constant, Ka , for the reaction above.
Ka =
[H 3O + ][C9 H 7O4 - ]
[HC9 H7O4 ]
1 point is earned for the correct expression.
(b) Calculate the value of Ka for acetylsalicylic acid.
[H3O+] = 10−pH = 10−2.78 = 1.66 × 10−3 M
[H3O+] = [C9H7O4−] = 1.66 × 10−3 M
1 point is earned for the correct [H3O+].
[HC9H7O4] = 0.0100 M − 1.66 × 10−3 M
Ka =
-3 2
(1.66 × 10 )
= 3.3 × 10-4
0.0100 - (1.66 × 10-3 )
1 point is earned for the value of Ka .
(c) An aqueous solution of aspirin is buffered to have equal concentrations of HC9H7O4(aq) and
C9H7O4−(aq). Calculate the pH of the solution.
pH = pKa + log
[C 9 H 7 O 4 - ]
[ HC 9 H 7 O 4 ]
= −log(3.3 × 10−4 ) + 0
1 point is earned for a pH consistent with the Ka calculated in
part (b).
= 3.48
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Scoring Worksheet
The following provides a scoring worksheet and conversion table
used for calculating a composite score of the exam.
2015 AP Chemistry Scoring Worksheet
Section I: Multiple Choice
Number Correct
(out of 50)
× 1.0000 =
Weighted Section I Score
(Do not round)
Section II: Free Response
Question 1
Question 2
Question 3
(out of 10)
(out of 10)
(out of 10)
Question 4
Question 5
Question 6
Question 7
(out of 4)
(out of 4)
(out of 4)
(out of 4)
× 1.0869 =
(Do not round)
× 1.0869 =
(Do not round)
× 1.0869 =
(Do not round)
× 1.0869 =
(Do not round)
× 1.0869 =
(Do not round)
× 1.0869 =
(Do not round)
× 1.0869 =
Sum =
(Do not round)
Weighted
Section II
Score
(Do not round)
Composite Score
Weighted
Section I Score
+
Weighted
Section II Score
=
Composite Score
(Round to nearest
whole number)
AP Score Conversion Chart
Chemistry
Composite
Score Range
AP Score
5
69-100
4
53-68
3
36-52
2
25-35
1
0-24
AP Chemistry
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